Power of a power with integer and variable bases
We introduced the power of a power when we learnt about the zero index.
Consider the expression $\left(5^2\right)^3$(52)3. What is the resulting power of base $5$5? To find out, have a look at the expanded form of the expression:
| $\left(5^2\right)^3$(52)3 | $=$= | $\left(5\times5\right)^3$(5×5)3 |
| $=$= | $\left(5\times5\right)\times\left(5\times5\right)\times\left(5\times5\right)$(5×5)×(5×5)×(5×5) |
In the expanded form, we can see that $5$5 is multiplied by itself $6$6 times. That is $\left(5\times5\right)\times\left(5\times5\right)\times\left(5\times5\right)=5^6$(5×5)×(5×5)×(5×5)=56.
To avoid having to write every power of a power expression in expanded form, we want a shortcut in simplifying $\left(5^2\right)^3$(52)3 to get $5^6$56. The shortcut? We MULTIPLY the powers.
Generalising this, we get:
$\left(x^a\right)^b=x^{a\times b}$(xa)b=xa×b
Examples
Question 1
Express in simplified index form: $\left(3^5\right)^3$(35)3
Solution:
| $\left(3^5\right)^3$(35)3 | $=$= | $3^{5\times3}$35×3 |
| $=$= | $3^{15}$315 |
Question 2
Simplify using the index laws:
$\left(3^5\right)^3\times\left(3^2\right)^3$(35)3×(32)3
Question 3
Simplify and evaluate using the index laws: $\left(4^2\right)^3\div\left(4^4\right)^0$(42)3÷(44)0
Think: We will have to use the Power of a Power, Zero Index and Division Laws.
Starting with the numerator
| $\left(4^2\right)^3$(42)3 | $=$= | $4^{2\times3}$42×3 |
| $=$= | $4^6$46 |
Now the denominator
| $\left(4^4\right)^0$(44)0 | $=$= | $4^{4\times0}$44×0 |
| $=$= | $4^0$40 | |
| $=$= | $1$1 |
Combining the two:
| $\left(4^2\right)^3\div\left(4^4\right)^0$(42)3÷(44)0 | $=$= | $\frac{4^6}{1}$461 |
| $=$= | $4^6$46 |